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Solutions pg 453
Solution - Solution - homogeneous mixture of pure substances.
Solvent Solvent – Medium used to dissolve, present in greater amounts in mixture
Solute Solute - substance being dissolved
B. Solvation
Solvation – Solvation – the process of dissolving
solute particles are separated and pulled into solution
solute particles are surrounded by solvent particles
ent particles
The process of dissolution is favored by:
1) A decrease in the energy of the system (exothermic)
2) An increase in the disorder
of the system (entropy)
Liquids Dissolving in Liquids
Liquids that are soluble in one another (“mix”) are MISCIBLE.– “LIKE dissolves LIKE”
POLAR liquids are generally soluble in other POLAR liquids.
NONPOLAR liquids are generally soluble in other NONPOLAR liquids.
Factors affecting rate of dissolution: think iced tea vs. hot tea &
the type of sugar you use: cubes or granulated
1) Surface area / particle size– Greater surface area, faster it dissolves
2) Temperature– Most solids dissolve faster @ higher temps
3) Agitation– Stirring/shaking will speed up dissolution
Solution Concentration: pg. 462
Concentration refers to the amount of solute dissolved in a solution.
If something is a concentrate we usually say it is dissolved into something else
Ex. Orange juice concentrate. We mix it with water to drink it!
Saturation: a solid solute dissolves in a solvent until the soln is SATURATED
Unsaturated solution – is able to dissolve more solute
Saturated solution – has dissolved the maximum amount of solute
Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid crystals generally form when this solution is cooled.
C. Solubility
SATURATED SOLUTIONno more solute dissolves
UNSATURATED SOLUTIONmore solute dissolves
SUPERSATURATED SOLUTIONbecomes unstable, crystals form
concentration
SOLUBILITY
Solubility = the amount of solute that will dissolve in a given amount of solvent
We can use a solubility chart like on page 458 to figure this out!
Factors Affecting Solubility The nature of the solute and solvent:
different substances have different solubilities Temperature: many solids substances
become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps.
Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases. (Henry’s Law)
Solids are more soluble at...Solids are more soluble at...– high temperatures.
Gases are more soluble at...Gases are more soluble at...• low temperatures &• high pressures
(Henry’s Law).• EX: nitrogen narcosis,
the “bends,” soda
Gases: solubility Temp and Pressure (think: flat soda) Have you ever seen mentos in Diet Coke?
– Nucleation site: the following factors that contribute to the bubble formation:
Diet coke– carbon dioxide is what makes the
bubbles form in the first place – in synthetic mixtures aspartam, caffeine
and potassium benzoate where shown give better fountains
Mentos– the most important property is the rough surface which provides
plenty of nucleation sites for bubble formation – the density makes them sink which is ideal as the bubbles formed
at the bottom of the bottle help expel much more soda – mentos contains gelatin and gum arabic which could also reduce
surface tension
Molarity Calculations
molar mass(g/mol)
6.02 1023
(particles/mol)
MASS
IN
GRAMS
MOLESNUMBER
OF
PARTICLES
LITERSOFSOLUTION
Molarity(mol/L)
B. Molarity Calculations How many grams of NaCl are
required to make 0.500L of 0.25M NaCl?
0.500 L 0.25 mol
1 L
= 7.3 g NaCl
58.44 g
1 mol
L 1
mol0.25 0.25M
B. Molarity Calculations Find the molarity of a 250 mL
solution containing 10.0 g of NaF.
10.0 g 1 mol
41.99 g = 0.238 mol NaF
0.238 mol
0.25 L M = = 0.95M NaF
L
molM
Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) 5.00 M Na2SO4
L) M)(2.50 665.0()M)(V (5.00 1
mL 333 L 333.0V1
2211 VMVM
Add 0.333 L of Na2SO4 to 2.17 L of water.
Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:solid Na2SO4.
L 2.50
xM 665.0
nsol' L
solute mol(M)Molarity
mol 1.6625x
g236mol 1
g 04.142SONa mol 6625.1 42
Dissolve 236 g of Na2SO4 in enough water
to create 2.50 Lof solution.
MASS PERCENT
Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?
100g 176
g 24%14%6.13
MOLALITY Example: What is the molality of a
solution that contains 12.8 g of C6H12O6 in 187.5 g water?
solvent kg
solute molm
g 180.18
mol 1OHC g8.12 6126 mol 07104.0
kg 1875.0
mol 0.07104m m 379.0
MOLALITY Example: How many grams of H2O must be
used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C12H22O11?
solvent kg
solute molm
g 342.34
mol 1OHC g 50.0 112212 mol 0.1461
x
mol 0.1461m 1.25
OH 117gkg 1168.0 2x
Colligative Properties of Solutions (page 471)
Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.
Colligative Properties
Lowering vapor pressure Raising boiling point Lowering freezing point Generating an osmotic pressure
Boiling Point Elevation
a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.
Like when adding salt to a pot of boiling water to make pasta
Boiling Point Elevation
Tb = kbmwhere: Tb = elevation of boiling pt
m = molality of solute (mol solute/kg solvent)
kb = the molal boiling pt elevation constant
kb values are constants; see table 15.4 pg. 472
kb for water = 0.52 °C/m
Ex: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution?
“normal” implies 1 atm of pressure Tb = kbm
Tb = (0.52 C/m)(2.50 m)
Tb = 1.3 C
Tb = 100.0 C + 1.3 C = 101.3 C
Freezing/Melting Point Depression
The freezing point of a solution is always lower than that of the pure solvent.
Like when salting roads in snowy places so the roads don’t ice over or when making ice cream
Freezing/Melting Point Depression
Tf = kfmwhere: Tf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant
kf for water = 1.86 °C/m kf values are constants;
see table 15.4 pg. 472
Ex: Calculate the freezing pt of a 2.50 m glucose solution.
Tf = kfm
Tf = (1.86 C/m)(2.50 m)
Tf = 4.65 C
Tf = 0.00C - 4.65 C = -4.65C
Example
Calculate the freezing-point depression (ΔTf) of a benzene solution containing 400. g of benzene and 200. g of acetone, C3H6O (solute).
Kf for benzene is 5.12 °C/m
Answer
ΔTf = Kf x m
ΔTf = (5.12 °C/m) x (m)
m =1
200. 58.09
8.61 0.400
molg x
mol solute gm
kg solvent kg
So, ΔTf = (5.12 °C/m) x (8.61 m) = 44.1 °C
Ex: When 15.0 g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine Kf for formic acid.
Tf = kfm
1.20 C= (kf)( 0.4340 m)
kf = 2.76 C/m
mol 3255.0g 46.08
mol 1OHHC g 0.15 52
m 4340.0kg 75.0
mol 0.3255
Tf = kfm
Electrolytes and Colligative Properties• Colligative properties depend on the # of particles present in solution.• Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.
Electrolytes and Colligative Properties For example, the freezing pt of water is
lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m.– Such as C6H12O6, or any other covalent
compound
However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute.– NaCl Na+ + Cl- (2 particles)
Electrolytes - Boiling Point Elevation and Freezing Point Depression
The relationships are given by the following equations:
Tf = kf ·m·n or Tb = kb·m·n
Tf/b = f.p. depression/elevation of b.p.m = molality of solute
kf/b = b.p. elevation/f.p depression constantn = # particles formed from the dissociation of
each formula unit of the solute
Ex: What is the freezing pt of a 1.15 m sodium chloride solution?
NaCl Na+ + Cl- n=2
Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(2)
Tf = 4.28 C
Tf = 0.00C - 4.28 C = -4.28C